Problem: Suppose we wanted to evaluate the double integral $S = \iint_D xy \, dx \, dy$ by first applying a change of variables from $D$ to $R$ : $\begin{aligned} x &= X_1(u, v) = \dfrac{1}{2} \cos(2u) + v \\ \\ y &= X_2(u, v) = \dfrac{1}{2} \sin(2v) - u \end{aligned}$ What is $S$ under the change of variables? If you know an expression within absolute value is non-negative, do not use absolute value at all. $S = \iint_R $ $du \, dv$
If we have a transformation $\bold{X} : R \to D$, then we can rewrite an integral under the change of variables: $ \iint_D f(x, y) \, dA = \iint_R f(\bold{X}(u, v)) | J(\bold{X}) | \, du \, dv$ First we need to find the absolute value of the Jacobian, $|J(\bold{X})|$. $\begin{aligned} |J(\bold{X})| &= \left| \det \begin{pmatrix} \dfrac{\partial X_1}{\partial u} & \dfrac{\partial X_1}{\partial v} \\ \\ \dfrac{\partial X_2}{\partial u} & \dfrac{\partial X_2}{\partial v} \end{pmatrix} \right| \\ \\ &= \left| \det \begin{pmatrix} -\sin(2u) & 1 \\ \\ -1 & \cos(2v) \end{pmatrix} \right| \\ \\ &= \left| -\sin(2u)\cos(2v) + 1 \right| \\ \\ &= -\sin(2u)\cos(2v) + 1 \end{aligned}$ We can remove the absolute value signs in the final step because $-1 \leq \sin(2u)\cos(2v) \leq 1$ for any $u$ and $v$ we pick. As a result, the product plus one is always at least zero, so the absolute value has no effect. Now we substitute $u$ and $v$ in $f(x, y)$. $\begin{aligned} f(x, y) &= f(\bold{X}(u, v)) \\ \\ &= X_1(u, v)X_2(u, v) \\ \\ &= \left( \dfrac{1}{2}\cos(2u) + v \right) \left( \dfrac{1}{2} \sin(2v) - u\right) \end{aligned}$ Putting everything together, we get the integral under the change of variables: $ \iint_R (-\sin(2u)\cos(2v) + 1) \left( \dfrac{1}{2}\cos(2u) + v \right) \left( \dfrac{1}{2} \sin(2v) - u \right) \, du \, dv$